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Both are correct: for $\ce{Cl-}$, the inner set of electrons would be implicit, as $\ce{Cl}$ has 17 electrons. The point of drawing these orbital diagrams, especially in a molecular context, is to show the nature of bonds and to show how the octet/duplet rules are being followed. If you have difficulties in using the second diagram (finding it harder to ...


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.Hey, Araj! Welcome to ChemSE! Thanks for posting your question here. As far as the acidity of water, It is a very special case that observes the process of homoassociation. This is when an acid forms stabilizing hydrogen-bonds with it's conjugate base This allows water to be the more stable acid, because hydrogen bonds are observed between $\ce{OH-}$ and $\...


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(https://i.stack.imgur.com/znrms.jpg) we cannot determine the extent of s p and d character in hybrid orbitals. but we can say that s character is greater in axial positions as angle is more than in equatorial positions.


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The 'project' (which I suspect relates to the searching of large chemical databases with artificial intelligence algorithms) sounds like an attempt to construct a decision tree with associated probabilities at leaf points to which an information theory based entropy function is calculated to arrive at an optimal decision path to identify the chemical ...


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Nobody can state that a chemical structure does exist or does not exist. Experience is the only criteria. Let me give you a simple example, Chlorine and Bromine are supposed to be interchangeable. If a structure exist with Cl, usually that the corresponding structure exist with Br instead of Cl. It works usually. But look ! The following series of compounds ...


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I agree in part with Mithoron and MaxW's comments: these are different molecules, not even sharing a mutual atom (e.g. C-O vs C-C), so direct comparison is restricted. However, Pauling's concept of electronegativity does explain why, in general, the heteronuclear A-B bond is stronger than the average of the homonuclear A-A and B-B bonds. the difference in ...


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This question has a few different parts to it. Part 1 - what's that peak at 3.6ppm? Well, it's not the -OH peak. The peak at 3.6 is from the CH(OH). Part 2 - where is the -OH peak if that isn't it at 3.6ppm? Peaks from labile protons are frequently not observed, and have variable chemical shifts, unless great care is taken in sample preparation. This ...


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The unsaturated regions are electron-rich and typically allow for electrophilic attack, such as by halogens or protons. Saturated hydrocarbons usually react only with radicals (radical halogenation). C-H activation is often hard in saturated hydrocarbons. While double and triple bonds are stronger than a single bond when looking at complete cleavage of a ...


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When a substance is made of molecules, these molecules are made of atoms being strongly attached to one another. But a given molecule is not attached to its neighbor by strong bonds. On the contrary, they are attached by so-called van der Waals forces, which are rather weak. If you heat such a substance, the van der Waals forces are easy to destroy, and the ...


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They're for the entire bond (so for a triple bond, all 3). From Wikipedia, it can be defined as The standard enthalpy change when [a bond, be it single, double, or triple] is cleaved by homolysis to give fragments A and B, which are usually radical species So you get two fragments as your final state.


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Your $\pu{640 kJ mol^-1}$ value is correct, so the next step is to neutralize the ions. First add an electron to a potassium ion to get a potassium atom. This releases $\pu{418 kJ mol^-1}$ because it is the reverse of the potassium atom ionization. Then take away an electron from fluoride ion to get a fluorine atom. This requires an input of $\pu{328 kJ mol^-...


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The equation is $$\ce{K2CrO4 + Ba(NO3)2 -> BaCrO4 + 2 KNO3}$$ The acetic acid does not intervene. $\ce{BaCrO4}$ is insoluble in the solution and make a pale yellow precipitate, that can be filtrated.


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